Estimating $\sin\!\!\!\left(\frac{13}{12}\pi\right)$ using a Taylor polynomial about $x=\pi$, what is the least degree of the polynomial that assures an error smaller than $0.001$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $4$ (Choice C) C $5$ (Choice D) D $6$
Answer: We will use the Lagrange error bound. Let's assume the polynomial's degree is $n$. The absolute value of the $(n+1)^{\text{th}}$ derivative of $\sin(x)$ is bounded by $1$. [Why?] The Lagrange bound for the error assures that: $\left|R_n\left(\dfrac{13}{12}\pi\right) \right|\leq \dfrac{1}{(n+1)!}\left(\dfrac{13}{12}\pi-\pi \right)^{n+1}$ Solving $\dfrac{1}{(n+1)!}\left(\dfrac{\pi}{12}\right)^{n+1}<0.001$ using trial and error, we find that $n\geq3$. In conclusion, the least degree of the polynomial that assures our error bound is $3$.